Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $t \neq 0$. $q = \dfrac{t + 5}{5t^2 - 15t - 200} \div \dfrac{-t + 6}{2t^2 - 32t + 128} $
Explanation: Dividing by an expression is the same as multiplying by its inverse. $q = \dfrac{t + 5}{5t^2 - 15t - 200} \times \dfrac{2t^2 - 32t + 128}{-t + 6} $ First factor out any common factors. $q = \dfrac{t + 5}{5(t^2 - 3t - 40)} \times \dfrac{2(t^2 - 16t + 64)}{-(t - 6)} $ Then factor the quadratic expressions. $q = \dfrac {t + 5} {5(t - 8)(t + 5)} \times \dfrac {2(t - 8)(t - 8)} {-(t - 6)} $ Then multiply the two numerators and multiply the two denominators. $q = \dfrac {(t + 5) \times 2(t - 8)(t - 8) } { 5(t - 8)(t + 5) \times -(t - 6)} $ $q = \dfrac {2(t - 8)(t - 8)(t + 5)} {-5(t - 8)(t + 5)(t - 6)} $ Notice that $(t - 8)$ and $(t + 5)$ appear in both the numerator and denominator so we can cancel them. $q = \dfrac {2\cancel{(t - 8)}(t - 8)(t + 5)} {-5\cancel{(t - 8)}(t + 5)(t - 6)} $ We are dividing by $t - 8$ , so $t - 8 \neq 0$ Therefore, $t \neq 8$ $q = \dfrac {2\cancel{(t - 8)}(t - 8)\cancel{(t + 5)}} {-5\cancel{(t - 8)}\cancel{(t + 5)}(t - 6)} $ We are dividing by $t + 5$ , so $t + 5 \neq 0$ Therefore, $t \neq -5$ $q = \dfrac {2(t - 8)} {-5(t - 6)} $ $ q = \dfrac{-2(t - 8)}{5(t - 6)}; t \neq 8; t \neq -5 $